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Let A and B be points with position vect...

Let `A` and `B` be points with position vectors `veca` and `vecb` with respect to origin `O`. If the point `C` on `OA` is such that `2vec(AC)=vec(CO), vec(CD) ` is parallel to `vec(OB)` and `|vec(CD)|=3|vec(OB)|` then `vec(AD)` is (A) `vecb-veca/9` (B) `3vecb-veca/3` (C) `vecb-veca/3` (D) `vecb+veca/3`

A

`3b-(a)/(2)`

B

`3b+(a)/(2)`

C

`3b-(a)/(3)`

D

`3b+(a)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
Since, `OA=a,OB=b and 2AC=CO`
by section formula, `OC=(2)/(3)a`
Therefore, `|CD|=3|OB|`
`impliesCD=3b`
`impliesOD=OC+CD=(2)/(3)a+3b`
Hence, `AD=OD-OA=(2)/(2)a+3b-a`
`=3a-(1)/(3)a`.
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